1. Given data for the misread observations

There are $n=15$ observations.

The mean and standard deviation computed from the wrong data are: $\bar{x}=8, \qquad \sigma=3$

So the variance of the wrong data is: $\sigma^2=9$

2. Find the wrong sum and wrong sum of squares

Using $\bar{x}=\frac{\sum x_i}{n}$ we get $\sum x_i = n\bar{x} = 15\cdot 8 = 120$

Also, $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$ So, $9 = \frac{\sum x_i^2}{15} - 8^2$ $9 = \frac{\sum x_i^2}{15} - 64$ $\frac{\sum x_i^2}{15} = 73$ $\sum x_i^2 = 15\cdot 73 = 1095$

Thus for the wrongly recorded data: $\sum x_i = 120, \qquad \sum x_i^2 = 1095$

3. Correct the misread value

One observation was actually $20$, but it was taken as $5$.

So the corrected sum is: $\sum x_i^{(c)} = 120 - 5 + 20 = 135$

The corrected sum of squares is: $\sum (x_i^{(c)})^2 = 1095 - 5^2 + 20^2$ $=1095-25+400=1470$

4. Find the correct mean

$\bar{x}_c = \frac{135}{15} = 9$

5. Find the correct variance

Using $\sigma_c^2 = \frac{\sum (x_i^{(c)})^2}{n} - \bar{x}_c^2$ we get $\sigma_c^2 = \frac{1470}{15} - 9^2$ $=98-81=17$

6. Final answer

The correct variance is: $\boxed{17}$

This matches the stored correct answer.